題目連結：

1110 Complete Binary Tree
Given a tree, you are supposed to tell if it is a complete binary tree.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤20) which is the total number of nodes in the tree – and hence the nodes are numbered from 0 to N−1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a - will be put at the position. Any pair of children are separated by a space.

Output Specification:
For each case, print in one line YES and the index of the last node if the tree is a complete binary tree, or NO and the index of the root if not. There must be exactly one space separating the word and the number.

Sample Input 1:

9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -

Sample Output 1:

YES 8

Sample Input 2:

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

Sample Output 2:

NO 1

思路：
判斷一個二叉樹是不是完整二叉樹，從根節點開始，依次放入佇列，如果左子樹或者右子樹為空，則記為-1，同時加入佇列。通過層次遍歷，到第一次出隊時為-1時結束，統計這時候已經便利了的節點數，如果等於總的節點數+1（因為包含了剛出隊為-1的節點），那麼則為完整二叉樹。否則不是完整二叉樹。

#include <iostream>
#include<cstdio>
#include<queue>
using
 
 namespace std;
struct node{
	int left,right;
}tree[25];
queue<int> Q; 
int in[25];
int root=0,last=-1,num=0;
void bfs(){
	Q.push(root);
	while(!Q.empty()){
		num++;
		int front=Q.front();
		Q.pop();
		if(front!=-1) last=front;
		else break;
		Q.push(tree[front].left);
		Q.push(tree[front].right);
	}
}
int main(int argc, char** argv) {
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		string l,r;
		cin>>l>>r;
		if(l[0]=='-') tree[i].left=-1;
		else{
			in[stoi(l)]++;
			tree[i].left=stoi(l);
		} 
		if(r[0]=='-') tree[i].right=-1;
		else{
			in[stoi(r)]++;
			tree[i].right=stoi(r);
		} 
	}
	while(in[root]) root++;
	bfs();
	if(num-1!=n){
		printf("NO %d\n",root);
	}else{
		printf("YES %d\n",last);
	}
	return 0;
}